高橋君とカード / Tak and Cards AtCoder - 2037 （DP）

Problem Statement

 

Tak has N cards. On the i-th (1≤i≤N) card is written an integer x_i. He is selecting one or more cards from these N cards, so that the average of the integers written on the selected cards is exactly A. In how many ways can he make his selection?

Constraints

 

• 1≤N≤50
• 1≤A≤50
• 1≤x_i≤50
• N, A, x_i are integers.

Partial Score

 

• 200 points will be awarded for passing the test set satisfying 1≤N≤16.

Input

 

The input is given from Standard Input in the following format:

N A
x1 x2 … xN

Output

 

Print the number of ways to select cards such that the average of the written integers is exactly A.

Sample Input 1

 

4 8
7 9 8 9

Sample Output 1

 

5

• The following are the 5 ways to select cards such that the average is 8:
  • Select the 3-rd card.
  • Select the 1-st and 2-nd cards.
  • Select the 1-st and 4-th cards.
  • Select the 1-st, 2-nd and 3-rd cards.
  • Select the 1-st, 3-rd and 4-th cards.

Sample Input 2

 

3 8
6 6 9

Sample Output 2

 

0

Sample Input 3

 

8 5
3 6 2 8 7 6 5 9

Sample Output 3

 

19

Sample Input 4

 

33 3
3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3

Sample Output 4

 

8589934591

• The answer may not fit into a 32-bit integer.

 

题意：

给了你N个数，和一个数a，

让你从这N个数中选择出一些数，并使这些数的sum和是a的倍数。

问你有多少种方案数。

思路：

我们定义dp[i][j] 代表选了i个数时，sum和为j的方案数。

初始化dp[0][0]=1

转移方程时dp[i+1][j+x]+=dp[i][j]，x为输入的数。

细节见代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <vector>
#include <iomanip>
#define ALL(x) (x).begin(), (x).end()
#define rt return
#define dll(x) scanf("%I64d",&x)
#define xll(x) printf("%I64d\n",x)
#define sz(a) int(a.size())
#define all(a) a.begin(), a.end()
#define rep(i,x,n) for(int i=x;i<n;i++)
#define repd(i,x,n) for(int i=x;i<=n;i++)
#define pii pair<int,int>
#define pll pair<long long ,long long>
#define gbtb ios::sync_with_stdio(false),cin.tie(0),cout.tie(0)
#define MS0(X) memset((X), 0, sizeof((X)))
#define MSC0(X) memset((X), '\0', sizeof((X)))
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define eps 1e-6
#define gg(x) getInt(&x)
#define db(x) cout<<"== [ "<<x<<" ] =="<<endl;
using namespace std;
typedef long long ll;
ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
ll lcm(ll a,ll b){return a/gcd(a,b)*b;}
ll powmod(ll a,ll b,ll MOD){ll ans=1;while(b){if(b%2)ans=ans*a%MOD;a=a*a%MOD;b/=2;}return ans;}
inline void getInt(int* p);
const int maxn=1000010;
const int inf=0x3f3f3f3f;
/*** TEMPLATE CODE * * STARTS HERE ***/
ll dp[55][55*55];
int k;
int n;
int main()
{
    //freopen("D:\\common_text\\code_stream\\in.txt","r",stdin);
    //freopen("D:\\common_text\\code_stream\\out.txt","w",stdout);
    gbtb;
    cin>>n>>k;
    int x;
    dp[0][0]=1ll;
    repd(i,1,n)
    {
        cin>>x;
        for(int j=i-1;j>=0;j--)// 枚举i-1~0, 一定倒序枚举，以避免重复相加。
        {
            for(int w=0;w<=55*j;w++)// 枚举选了j个数的所有可能得到的值。
            {
                dp[j+1][w+x]+=dp[j][w]; // 加上贡献。
            }
        }
    }
    ll ans=0ll;
    repd(i,1,n)
    {
        ans+=dp[i][i*k];// 根据平均数的公式计算答案。
    }
    cout<<ans<<endl;


    return 0;
}

inline void getInt(int* p) {
    char ch;
    do {
        ch = getchar();
    } while (ch == ' ' || ch == '\n');
    if (ch == '-') {
        *p = -(getchar() - '0');
        while ((ch = getchar()) >= '0' && ch <= '9') {
            *p = *p * 10 - ch + '0';
        }
    }
    else {
        *p = ch - '0';
        while ((ch = getchar()) >= '0' && ch <= '9') {
            *p = *p * 10 + ch - '0';
        }
    }
}